Chapter 4: The Mole and Stoichiometry

1
Chapter 4 The Mole and Stoichiometry

• Chemistry The Molecular Nature of Matter, 6E
• Jespersen/Brady/Hyslop

2
Stoichiometry

• Mass balance of all formulas involved in chemical
  reactions
• Stoichiometric Calculations
• Conversions from one set of units to another
  using Dimensional Analysis
• Need to know
• Equalities to make conversion factors
• Steps to go from starting units to desired units

3
Using Mass to Count
4
Molecular to Laboratory Scale

• So far, we have looked at chemical formulas
  reactions at a molecular scale.
• It is known from experiments that
• Electrons, neutrons protons have set masses.
• Atoms must also have characteristic masses
• Just extremely small
• Need a way to scale up chemical formulas
  reactions to carry out experiments in laboratory
• Mole is our conversion factor

5
The Mole

• Number of atoms in exactly 12 grams of 12C atoms
• How many atoms in 1 mole of 12C ?
• Based on experimental evidence
• 1 mole of 12C 6.022 1023 atoms 12.011 g
• Avogadros number NA
• Number of atoms, molecules or particles in one
  mole
• 1 mole of X 6.022 1023 units of X
• 1 mole Xe 6.0221023 Xe atoms
• 1 mole NO2 6.0221023 NO2 molecules

6
Moles of Compounds

• Atoms
• Atomic Mass
• Mass of atom (from periodic table)
• 1 mole of atoms gram atomic mass
  6.0221023 atoms
• Molecules
• Molecular Mass
• Sum of atomic masses of all atoms in compounds
  formula
• 1 mole of molecule X gram molecular mass of X
• 6.022 1023 molecules

7
Moles of Compounds

• Ionic compounds
• Formula Mass
• Sum of atomic masses of all atoms in ionic
  compounds formula
• 1 mole ionic compound X gram formula mass of X
• 6.022 1023 formula units
• General
• Molar mass (MM)
• Mass of 1 mole of substance (element, molecule,
  or ionic compound) under consideration
• 1 mol of X gram molar mass of X 6.022
  1023 formula units

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SI Unit for Amount Mole

• 1 mole of substance X gram molar mass of X
• 1 mole S 32.06 g S
• 1 mole NO2 46.01 g NO2
• Molar mass is our conversion factor between g
  moles
• 1 mole of X 6.022 1023 units of X
• NA is our conversion factor between moles
  molecules
• 1 mole H2O 6.022 1023 molecules H2O
• 1 mole NaCl 6.022 1023 formula units NaCl

9
Learning Check Using Molar Mass

• Ex. How many moles of iron (Fe) are in 15.34 g
  Fe?
• What do we know?
• 1 mol Fe 55.85 g Fe
• What do we want to determine?
• 15.34 g Fe ? Mol Fe
• Set up ratio so that what you want is on top
  what you start with is on the bottom

0.2747 mole Fe
10
Learning Check Using Molar Mass

• Ex. If we need 0.168 mole Ca3(PO4)2 for an
  experiment, how many grams do we need to weigh
  out?
• Calculate MM of Ca3(PO4)2
• 3 mass Ca 3 40.08 g 120.24 g
• 2 mass P 2 30.97 g 61.94 g
• 8 mass O 8 16.00 g 128.00 g
• 1 mole Ca3(PO4)2 310.18 g Ca3(PO4)2
• What do we want to determine?
• 0.168 g Ca3(PO4)2 ? Mol Fe

11
Learning Check Using Molar Mass

• Set up ratio so that what you want is on the top
  what you start with is on the bottom

52.11 g Ca3(PO4)2
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Your Turn!

• How many moles of CO2 are there in 10.0 g?
• 1.00 mol
• 0.0227 mol
• 4.401 mol
• 44.01 mol
• 0.227 mol

0.227 mol CO2
13
Your Turn!

• How many grams of platinum (Pt) are in 0.475 mole
  Pt?
• 195 g
• 0.0108 g
• 0.000513 g
• 0.00243 g
• 92.7 g

Molar mass of Pt 195.08 g/mol
92.7 g Pt
14
Using Moles in Calculations

• Start with either
• Grams (Macroscopic)
• Elementary units (Microscopic)
• Use molar mass to convert from grams to mole
• Use Avogadros number to convert from moles to
  elementary units

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Macroscopic to Microscopic

• How many silver atoms are in a 85.0 g silver
  bracelet?
• What do we know?
• 107.87 g Ag 1 mol Ag
• 1 mol Ag 6.0221023 Ag atoms
• What do we want to determine?
• 85.0 g silver ? atoms silver
• g Ag ?? mol Ag ?? atoms Ag
• 4.7 1023 Ag atoms

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Using Avogadros Number

• What is the mass, in grams, of one molecule of
  octane, C8H18?
• Molecules octane ?? mol octane ?? g octane
• 1. Calculate molar mass of octane
• Mass C 8 12.01 g 96.08 g
• Mass H 18 1.008 g 18.14 g
• 1 mol octane 114.22 g octane
• 2. Convert 1 molecule of octane to grams
• 1.897 1022 g octane

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Learning Check Mole Conversions

• Calculate the number of formula units of Na2CO3
  in 1.29 moles of Na2CO3.
• How many moles of Na2CO3 are there in 1.15 x 105
  formula units of Na2CO3 ?

7.771023 particles Na2CO3
1.911019 mol Na2CO3
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Your Turn!

• How many atoms are in 1.00 x 109 g of U (1 ng)?
  Molar mass U 238.03 g/mole.
• 6.02 x 1014 atoms
• 4.20 x 1011 atoms
• 2.53 x 1012 atoms
• 3.95 x 1031 atoms
• 2.54 x 1021 atoms

2.53 x 1012 atoms U
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Your Turn!

• Calculate the mass in grams of FeCl3 in 1.53
  1023 formula units. (molar mass 162.204 g/mol)
• 162.2 g
• 0.254 g
• 1.6611022 g
• 41.2 g
• 2.37 1022

41.2 g FeCl3
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Mole-to-Mole Conversion Factors

• Can use chemical formula to relate amount of each
  atom to amount of compound
• In H2O there are 3 relationships
• 2 mol H ? 1 mol H2O
• 1 mol O ? 1 mol H2O
• 2 mol H ? 1 mol O
• Can also use these on atomic scale
• 2 atom H ? 1 molecule H2O
• 1 atom O ? 1 molecule H2O
• 2 atom H ? 1 molecule O

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Stoichiometric Equivalencies

• Within chemical compounds, moles of atoms always
  combine in the same ratio as the individual atoms
  themselves
• Ratios of atoms in chemical formulas must be
  whole numbers!!
• These ratios allow us to convert between moles of
  each quantity
• Ex. N2O5
• 2 mol N ? 1 mol N2O5
• 5 mol O ? 1 mol N2O5
• 2 mol N ? 5 mol O

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Stoichiometric Equivalencies
Equivalency Mole Ratio Mole Ratio
2 mol N ? 1 mol N2O5 2 mol N 1 mol N2O5
2 mol N ? 1 mol N2O5 1 mol N2O5 2 mol N
5 mol O ? 1 mol N2O5 5 mol O 1 mol N2O5
5 mol O ? 1 mol N2O5 1 mol N2O5 5 mol O
2 mol N ? 5 mol O 5 mol O 2 mol N
2 mol N ? 5 mol O 2 mol N 5 mol O
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Calculating the Amount of a Compound by Analyzing
One Element

• Calcium phosphate is widely found in natural
  minerals, bones, and some kidney stones. A sample
  is found to contain 0.864 moles of phosphorus.
  How many moles of Ca3(PO4)2 are in that sample?
• What do we want to find?
• 0.864 mol P ? mol Ca3(PO4)2
• What do we know?
• 2 mol P ? 1 mol Ca3(PO4)2
• Solution

0.432 mol Ca3(PO4)2
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Your Turn!

• Calculate the number of moles of calcium in 2.53
  moles of Ca3(PO4)2
• 2.53 mol Ca
• 0.432 mol Ca
• 3.00 mol Ca
• 7.59 mol Ca
• 0.843 mol Ca
• 2.53 moles of Ca3(PO4)2 ? mol Ca
• 3 mol Ca ? 1 mol Ca3(PO4)2

7.59 mol Ca
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Mass-to-Mass Calculations

• Common laboratory calculation
• Need to know what mass of reagent B is necessary
  to completely react given mass of reagent A to
  form a compound
• Stoichiometry comes from chemical formula of
  compounds
• Subscripts
• Summary of steps
• mass A ? moles A ? moles B ? mass B

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Mass-to-Mass Calculations

• Chlorophyll, the green pigment in leaves, has the
  formula C55H72MgN4O5. If 0.0011 g of Mg is
  available to a plant for chlorophyll synthesis,
  how many grams of carbon will be required to
  completely use up the magnesium?
• Analysis
• 0.0011 g Mg ? ? g C
• 0.0011 g Mg ? mol Mg ? mol C ? g C
• Assembling the tools
• 24.3050 g Mg 1 mol Mg
• 1 mol Mg ? 55 mol C
• 1 mol C 12.011 g C

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Ex. Mass-to-Mass Conversion
0.0011 g Mg ?? mol Mg ?? mol C ?? g C
0.030 g C
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Your Turn!

• How many g of iron are required to use up all of
  25.6 g of oxygen atoms (O) to form Fe2O3?
• 59.6 g
• 29.8 g
• 89.4 g
• 134 g
• 52.4 g
• 59.6 g Fe

mass O ? mol O ? mol Fe ? mass Fe
25.6 g O ? ? g Fe
3 mol O ? 2 mol Fe
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Percentage Composition

• Way to specify relative masses of each element in
  a compound
• List of percentage by mass of each element
• Percentage by Mass
• Ex. Na2CO3 is
• 43.38 Na
• 11.33 C
• 45.29 O
• What is sum of by mass?

100.00
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Ex. Percent Composition

• Determine percentage composition based on
  chemical analysis of substance
• Ex. A sample of a liquid with a mass of 8.657 g
  was decomposed into its elements and gave 5.217 g
  of carbon, 0.9620 g of hydrogen, and 2.478 g of
  oxygen. What is the percentage composition of
  this compound?
• Analysis
• Calculate by mass of each element in sample
• Tools
• Eqn for by mass
• Total mass 8.657 g
• Mass of each element

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Ex. Composition of Compound

• For C
• For H
• For O
• composition tells us mass of each element in
  100.00 g of substance
• In 100.00 g of our liquid
• 60.26 g C, 11.11 g H 28.62 g O

60.26 C
11.11 H
28.62 O
Sum of percentages 99.99
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Your Turn!

• A sample was analyzed and found to contain 0.1417
  g nitrogen and 0.4045 g oxygen. What is the
  percentage composition of this compound?
• 1. Calculate total mass of sample
• Total sample mass 0.1417 g 0.4045 g 0.5462
  g
• 2. Calculate Composition of N
• 3. Calculate Composition of O

25.94 N
74.06 O
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Percent Compositions Chemical Identity

• Theoretical or Calculated Composition
• Calculated from molecular or ionic formula.
• Lets you distinguish between multiple compounds
  formed from the same 2 elements
• If experimental percent composition is known
• Calculate Theoretical Composition from proposed
  Chemical Formula
• Compare with experimental composition
• Ex. N O form multiple compounds
• N2O, NO, NO2, N2O3, N2O4, N2O5

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Ex. Using Percent Composition

• Are the mass percentages 30.54 N 69.46 O
  consistent with the formula N2O4?
• Procedure
• Assume 1 mole of compound
• Subscripts tell how many moles of each element
  are present
• 2 mol N 4 mol O
• Use molar masses of elements to determine mass of
  each element in 1 mole
• Molar Mass of N2O4 92.14 g N2O4 / 1 mol
• Calculate by mass of each element

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Ex. Using Percent Composition (cont)
28.14 g N
64.00 g O
30.54 N in N2O4
69.46 N in N2O4

• The experimental values match the theoretical
  percentages for the formula N2O4.

36
Your Turn

• If a sample containing only phosphorous oxygen
  has percent composition 56.34 P 43.66 O, is
  this P4O10?
• Yes
• No

4 mol P ? 1 mol P4O10
10 mol O ? 1 mol P4O10
4 mol P 4 ? 30.97 g/mol P 123.9 g P
10 mol O 10 ?16.00 g/mol O 160.0 g O
1 mol P4O10 283.9 g P4O10
43.64 P
56.36 O
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Determining Empirical Molecular Formulas

• When making or isolating new compounds one must
  characterize them to determine structure
• Molecular Formula
• Exact composition of one molecule
• Exact whole ratio of atoms of each element in
  molecule
• Empirical Formula
• Simplest ratio of atoms of each element in
  compound
• Obtained from experimental analysis of compound

Empirical formula CH2O
glucose
Molecular formula C6H12O6
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Three Ways to Calculate Empirical Formulas

• From Masses of Elements
• Ex. 2.448 g sample of which 1.771 g is Fe and
  0.677 g is O.
• From Percentage Composition
• Ex. 43.64 P and 56.36 O.
• From Combustion Data
• Given masses of combustion products
• Ex. The combustion of a 5.217 g sample of a
  compound of C, H, and O in pure oxygen gave 7.406
  g CO2 and 4.512 g of H2O.

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Strategy for Determining Empirical Formulas

• Determine mass in g of each element
• Convert mass in g to moles
• Divide all quantities by smallest number of moles
  to get smallest ratio of moles
• Convert any non-integers into integer numbers.
• If number ends in decimal equivalent of fraction,
  multiply all quantities by least common
  denominator
• Otherwise, round numbers to nearest integers

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1. Empirical Formula from Mass Data

• When a 0.1156 g sample of a compound was
  analyzed, it was found to contain 0.04470 g of C,
  0.01875 g of H, and 0.05215 g of N. Calculate the
  empirical formula of this compound.

Step 1 Calculate moles of each substance
3.722 ? 10?3 mol C
1.860 ? 10?2 mol H
3.723 ? 10?3 mol N
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1. Empirical Formula from Mass Data

• Step 2 Select the smallest of moles.
• Lowest is 3.722 x 103 mole
• C
• H
• Step 3 Divide all of moles by the smallest one

Mole ratio
Integer ratio
1.000
1
5
4.997
1.000
1

• N

Empirical formula CH5N
42
Empirical Formula from Mass Composition

• One of the compounds of iron and oxygen, black
  iron oxide, occurs naturally in the mineral
  magnetite. When a 2.448 g sample was analyzed it
  was found to have 1.771 g of Fe and 0.677 g of O.
  Calculate the empirical formula of this compound.
• Assembling the tools
• 1 mol Fe 55.845 g Fe 1 mol O 16.00 g
  O
• 1. Calculate moles of each substance

0.03171 mol Fe
0.0423 mol O
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1. Empirical Formula from Mass Data

• 2. Divide both by smallest mol to get smallest
  whole ratio.

3 3.000 Fe
1.000 Fe
3 3.99 O
1.33 O
Or
Empirical Formula Fe3O4
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2. Empirical Formula from Composition

• New compounds are characterized by elemental
  analysis, from which the percentage composition
  can be obtained
• Use percentage composition data to calculate
  empirical formula
• Must convert composition to grams
• Assume 100.00 g sample
• Convenient
• Sum of composition 100
• Sum of masses of each element 100 g

45
2. Empirical Formula from Composition

• Calculate the empirical formula of a compound
  whose composition data is 43.64 P and 56.36
  O. If the molar mass is determined to be 283.9
  g/mol, what is the molecular formula?
• Step 1 Assume 100 g of compound.
• 43.64 g P
• 56.36 g O

1 mol P 30.97 g
1 mol O 16.00 g
1.409 mol P
3.523 mol P
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2. Empirical Formula from Composition

• Step 2 Divide by smallest number of moles

? 2 2
? 2 5
Step 3 Multiple by n to get smallest integer
ratio
Here n 2
Empirical formula P2O5
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3. Empirical Formulas from Indirect Analysis

• In practice, compounds are not broken down into
  elements, but are changed into other compounds
  whose formula is known.
• Combustion Analysis
• Compounds containing carbon, hydrogen, oxygen,
  can be burned completely in pure oxygen gas
• Only carbon dioxide water are produced
• Ex. Combustion of methanol (CH3OH)
• 2CH3OH 3O2 ?? 2CO2 4H2O

48
Combustion Analysis
Classic
Modern CHN analysis
49
3. Empirical Formulas from Indirect Analysis

• Carbon dioxide water separated weighed
  separately
• All C ends up as CO2
• All H ends up as H2O
• Mass of C can be derived from amount of CO2
• mass CO2 ? mol CO2 ? mol C ? mass C
• Mass of H can be derived from amount of H2O
• mass H2O ? mol H2O ? mol H ? mass H
• Mass of oxygen is obtained by difference
• mass O mass sample (mass C mass H)

50
Ex. Indirect or Combustion Analysis

• The combustion of a 5.217 g sample of a compound
  of C, H, and O in pure oxygen gave 7.406 g CO2
  and 4.512 g of H2O. Calculate the empirical
  formula of the compound.

C H O CO2
MM (g/mol) 12.011 1.008 15.999 44.01
1. Calculate mass of C from mass of CO2. mass CO2
? mole CO2 ? mole C ? mass C
2.021 g C
51
Ex. Indirect or Combustion Analysis

• The combustion of a 5.217 g sample of a compound
  of C, H, and O gave 7.406 g CO2 and 4.512 g of
  H2O. Calculate the empirical formula of the
  compound.

2. Calculate mass of H from mass of H2O. mass
H2O ? mol H2O ? mol H ? mass H
0.5049 g H
3. Calculate mass of O from difference.
5.217 g sample 2.021 g C 0.5049 g H
2.691 g O
52
Ex. Indirect or Combustion Analysis
C H O
MM 12.011 1.008 15.999
g
2.021
0.5049
2.691
4. Calculate mol of each element
0.1683 mol C
0.5009 mol H
0.1682 mol O
53
Ex. Indirect or Combustion Analysis

• Preliminary empirical formula
• C0.1683H0.5009O0.1682
• 5. Calculate mol ratio of each element
• Since all values are close to integers, round to

C1.00H2.97O1.00
Empirical Formula CH3O
54
Determining Molecular Formulas

• Empirical formula
• Accepted formula unit for ionic compounds
• Molecular formula
• Preferred for molecular compounds
• In some cases molecular empirical formulas are
  the same
• When they are different, the subscripts of
  molecular formula are integer multiples of those
  in empirical formula
• If empirical formula is AxBy
• Molecular formula will be AnxBny

55
Determining Molecular Formula

• Need molecular mass empirical formula
• Calculate ratio of molecular mass to mass
  predicted by empirical formula round to nearest
  integer
• Ex. Glucose
• Molecular mass is 180.16 g/mol
• Empirical formula CH2O
• Empirical formula mass 30.03 g/mol

Molecular formula C6H12O6
56
Learning Check

• The empirical formula of a compound containing
  phosphorous and oxygen was found to be P2O5. If
  the molar mass is determined to be 283.9 g/mol,
  what is the molecular formula?
• Step 1 Calculate empirical mass

Step 2 Calculate ratio of molecular to
empirical mass
2
Molecular formula P4O10
57
Your Turn!

• The empirical formula of hydrazine is NH2, and
  its molecular mass is 32.0. What is its molecular
  formula?
• NH2
• N2H4
• N3H6
• N4H8
• N1.5H3

Molar mass of NH2 (114.01)g (21.008)g
16.017g
n (32.0/16.02) 2
Atomic Mass N14.007 H1.008 O15.999
58
Balanced Chemical Equations

• Useful tool for problem solving
• Prediction of reactants and products
• All atoms present in reactants must also be
  present among products.
• Coefficients are multipliers that are used to
  balance equations
• Two step process
• Write unbalanced equation
• Given products reactants
• Organize with plus signs arrow
• Adjust coefficients to get equal numbers of each
  kind of atom on both sides of arrow.

59
Guidelines for Balancing Equations

• Start balancing with the most complicated formula
  first.
• Elements, particularly H2 O2, should be left
  until the end.
• Balance atoms that appear in only two formulas
  one as a reactant the other as a product.
• Leave elements that appear in three or more
  formulas until later.
• Balance as a group those polyatomic ions that
  appear unchanged on both sides of the arrow.

60
Balancing Equations

• Use the inspection method
• Step 1. Write unbalanced equation
• Zn(s) HCl(aq) ? ZnCl2(aq) H2(g)
  unbalanced
• Step 2. Adjust coefficients to balance numbers of
  each kind of atom on both sides of arrow.
• Since ZnCl2 has 2Cl on the product side, 2HCl on
  reactant side is needed to balance the equation.
• Zn(s) 2HCl(aq) ?? ZnCl2(aq) H2(g)
• 1 Zn each side
• 2 H each side
• So balanced

61
Learning Check Balancing Equations

• AgNO3(aq) Na3PO4(aq) ? Ag3PO4(s) NaNO3(aq)
• Count atoms
• Reactants Products
• 1 Ag 3 Ag
• 3 Na 1 Na
• Add in coefficients by multiplying Ag Na by 3
  to get 3 of each on both sides
• 3AgNO3(aq) Na3PO4(aq) ? Ag3PO4(s) 3NaNO3(aq)
• Now check polyatomic ions
• 3 NO3? 3 NO3?
• 1 PO43? 1 PO43?
• Balanced

62
Balance by Inspection

• __C3H8(g) __O2(g) ? __CO2(g) __H2O(l)
• Assume 1 in front of C3H8
• 3C 1C ? 3
• 8H 2H ? 4
• 1C3H8(g) __O2(g) ? 3CO2(g) 4H2O(l)
• 2O ? 5 10 O (3 ? 2) 4 10
• 8H H 2 ? 4 8
• 1C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(l)

63
Your Turn!
Balance each of the following equations. What
are the coefficients in front of each compound?
__ Ba(OH)2(aq) __ Na2SO4(aq) ? __ BaSO4(s) __
NaOH(aq)
1
1
2
1
2
2
3
___KClO3(s) ? ___KCl(s) ___ O2(g)
3
1
6
2
__H3PO4(aq) __ Ba(OH)2(aq) ? __Ba3(PO4)2(s)
__H2O(l)
64
Using Balanced Equations Reaction Stoichiometry

• Balanced equation
• Critical link between substances involved in
  chemical reactions
• Gives relationship between amounts of reactants
  used amounts of products likely to be formed
• Numeric coefficient tells us
• The mole ratios for reactions
• How many individual particles are needed in
  reaction on microscopic level
• How many moles are necessary on macroscopic level

65
Stoichiometric Ratios

• Consider the reaction
• N2 3H2 ? 2NH3
• Could be read as
• When 1 molecule of nitrogen reacts with 3
  molecules of hydrogen, 2 molecules of ammonia are
  formed.
• Molecular relationships
• 1 molecule N2 ? 2 molecule NH3
• 3 molecule H2 ? 2 molecule NH3
• 1 molecule N2 ? 3 molecule H2

66
Stoichiometric Ratios

• Consider the reaction
• N2 3H2 ? 2NH3
• Could also be read as
• When 1 mole of nitrogen reacts with 3 moles of
  hydrogen, 2 moles of ammonia are formed.
• Molar relationships
• 1 mole N2 ? 2 mole NH3
• 3 mole H2 ? 2 mole NH3
• 1 mole N2 ? 3 mole H2

67
Using Stoichiometric Ratios

• Ex. For the reaction N2 3 H2 ? 2NH3, how many
  moles of N2 are used when 2.3 moles of NH3 are
  produced?
• Assembling the tools
• 2 moles NH3 1 mole N2
• 2.3 mole NH3 ? moles N2

1.2 mol N2
68
Your Turn!

• If 0.575 mole of CO2 is produced by the
  combustion of propane, C3H8, how many moles of
  oxygen are consumed? The balanced equation is
• C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(g)
• 0.575 mole
• 2.88 mole
• 0.192 mole
• 0.958 mole
• 0.345 mole

• Assembling the tools
• 0.575 mole CO2 ? moles O2
• 3 moles CO2 5 mole O2

0.958 mol O2
69
Mass-to-Mass Conversions

• Most common stoichiometric conversions that
  chemists use involve converting mass of one
  substance to mass of another.
• Use molar mass A to convert grams A to moles A
• Use chemical equations to relate moles A to moles
  B
• Use molar mass B to convert to moles B to grams B

70
Using Balanced Equation to Determine
Stoichiometry

• Ex. What mass of O2 will react with 96.1 g of
  propane (C3H8) gas, to form gaseous carbon
  dioxide water?
• Strategy
• 1. Write the balanced equation
• C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(g)
• 2. Assemble the tools
• 96.1 g C3H8 ? moles C3H8 ? moles O2 ? g O2
• 1 mol C3H8 44.1 g C3H8
• 1 mol O2 32.00 g O2
• 1 mol C3H8 5 mol O2

71
Using Balanced Equation to Determine
Stoichiometry

• Ex. What mass of O2 will react with 96.1 g of
  propane in a complete combustion?
• C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(g)
• 3. Assemble conversions so units cancel correctly

349 g of O2 are needed
72
Your Turn!

• How many grams of Al2O3 are produced when 41.5 g
  Al react?
• 2Al(s) Fe2O3(s) ?? Al2O3(s) 2Fe(l)
• 78.4 g
• 157 g
• 314 g
• 22.0 g
• 11.0 g

78.4 g Al2O3
73
Molecular Level of Reactions

• Consider industrial synthesis of ethanol
• C2H4 H2O ?? C2H5OH
• 3 molecules ethylene 3 molecules water react to
  form 3 molecules ethanol

74
Molecular Level of Reactions

• What happens if these proportions are not met?
• 3 molecules ethylene 5 molecules of oxygen
• All ethylene will be consumed some oxygen will
  be left over

75
Limiting Reactant

• Reactant that is completely used up in the
  reaction
• Present in lower of moles
• It determines the amount of product produced
• For this reaction ethylene
• Excess reactant
• Reactant that has some amount left over at end
• Present in higher of moles
• For this reaction water

76
Limiting Reactant Calculations

• Write the balanced equation.
• Identify the limiting reagent.
• Calculate amount of reactant B needed to react
  with reactant B
• Compare amount of B you need with amount of B you
  actually have.
• If need more B than you have, then B is limiting
• If need less B than you have, then A is limiting

77
Limiting Reactant Calculations

• Calculate mass of desired product, using amount
  of limiting reactant mole ratios.

78
Ex. Limiting Reactant Calculation

• How many grams of NO can form when 30.0 g NH3 and
  40.0 g O2 react according to
• 4 NH3 5 O2 ?? 4 NO 6 H2O
• Solution Step 1
• mass NH3 ? mole NH3 ? mole O2 ? mass O2
• Assembling the tools
• 1 mol NH3 17.03 g
• 1 mol O2 32.00 g
• 4 mol NH3 ? 5 mol O2

Only have 40.0 g O2, O2 limiting reactant
70.5 g O2 needed
79
Ex. Limiting Reactant Calculation

• How many grams of NO can form when 30.0 g NH3 and
  40.0 g O2 react according to
• 4 NH3 5 O2 ?? 4 NO 6 H2O
• Solution Step 2
• mass O2 ? mole O2 ? mole NO ? mass NO
• Assembling the tools
• 1 mol O2 32.00 g
• 1 mol NO 30.01 g
• 5 mol O2 ? 4 mol NO

Can only form 30.0 g NO.
30.0 g NO formed
80
Your Turn!

• If 18.1 g NH3 is reacted with 90.4 g CuO, what is
  the maximum amount of Cu metal that can be
  formed?
• 2NH3(g) 3CuO(s) ? N2(g) 3Cu(s)
  3H2O(g)
• (MM) (17.03) (79.55) (28.01)
  (64.55) (18.02)
• (g/mol)
• 127 g
• 103 g
• 72.2 g
• 108 g
• 56.5 g

127 g CuO needed. Only have 90.4g so CuO limiting
72.2 g Cu can be formed
81
Reaction Yield

• In many experiments, the amount of product is
  less than expected
• Losses occur for several reasons
• Mechanical issues sticks to glassware
• Evaporation of volatile (low boiling) products.
• Some solid remains in solution
• Competing reactions formation of by-products.
• Main reaction
• 2 P(s) 3 Cl2(g) ? 2 PCl3(l)
• Competing reaction
• PCl3(l) Cl2(g) ? PCl5(s) By-product

82
Theoretical vs. Actual Yield

• Theoretical Yield
• Amount of product that must be obtained if no
  losses occur.
• Amount of product formed if all of limiting
  reagent is consumed.
• Actual Yield
• Amount of product that is actually isolated at
  end of reaction.
• Amount obtained experimentally
• How much is obtained in mass units or in moles.

83
Percentage Yield

• Useful to calculate yield.
• Percent yield
• Relates the actual yield to the theoretical yield
• It is calculated as
• Ex. If a cookie recipe predicts a yield of 36
  cookies and yet only 24 are obtained, what is the
  yield?

84
Ex. Percentage Yield Calculation

• When 18.1 g NH3 and 90.4 g CuO are reacted, the
  theoretical yield is 72.2 g Cu. The actual yield
  is 58.3 g Cu. What is the percent yield?
• 2NH3(g) 3CuO(s) ? N2(g) 3Cu(s) 3H2O(g)

80.7
85
Learning Check Percentage Yield

• A chemist set up a synthesis of solid phosphorus
  trichloride by mixing 12.0 g of solid phosphorus
  with 35.0 g chlorine gas and obtained 42.4 g of
  solid phosphorus trichloride. Calculate the
  percentage yield of this compound.
• Analysis
• Write balanced equation
• P(s) Cl2(g) ?? PCl3(s)

86
Learning Check Percentage Yield

• Assembling the Tools
• 1 mol P 30.97 g P
• 1 mol Cl2 70.90 g Cl2
• 3 mol Cl2 ? 2 mol P
• Solution
• Determine Limiting Reactant
• But you only have 35.0 g Cl2, so Cl2 is limiting
  reactant

41.2 g Cl2
87
Learning Check Percentage Yield

• Solution
• Determine Theoretical Yield
• Determine Percentage Yield
• Actual yield 42.4 g

45.2 g PCl3
93.8
88
Your Turn!

• When 6.40 g of CH3OH was mixed with 10.2 g of O2
  and ignited, 6.12 g of CO2 was obtained. What
  was the percentage yield of CO2?
• 2CH3OH 3O2 ?? 2CO2
  4H2O
• MM(g/mol) (32.04) (32.00) (44.01)
  (18.02)
• 6.12
• 8.79
• 100
• 142
• 69.6

9.59 g O2 needed CH3OH limiting
8.79 g CO2 in theory
89
Stoichiometry Summary